theorem
  x<>y implies (Ex(x,p)).y = Ex(x,p.y)
proof
  assume
A1: x<>y;
  thus (Ex(x,p)).y = ('not' All(x,'not' p)).y by QC_LANG2:def 5
    .= 'not'(All(x,'not' p).y) by CQC_LANG:19
    .= 'not'(All(x,('not' p).y)) by A1,CQC_LANG:25
    .= 'not' All(x,'not'(p.y)) by CQC_LANG:19
    .= Ex(x,p.y) by QC_LANG2:def 5;
end;
